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\(=\dfrac{\sin35}{\cos35}\cdot\cot35+\dfrac{\cos55}{\sin55}\cdot\cot35\)
\(=\cot35\cdot\left(\dfrac{\sin35}{\cos35}+\dfrac{\sin35}{\cos35}\right)\)
\(=\dfrac{\cos35}{\sin35}\cdot\dfrac{2\sin35}{\cos35}=2\)
\(\cos^21^o+\cos^289^o=\cos^21^o+\cos^2\left(90^o-1^o\right)=\cos^21^o+\sin^21^o=1\)
\(\cos^22^o+\cos^288^o=\cos^22^o+\cos^2\left(90^o-2^o\right)=\cos^22^o+\sin^22^o=1\)
.......
\(\cos^244^o+\cos^246^o=\cos^244^o+\cos^2\left(90^o-44^o\right)=\cos^244^o+\sin^244^o=1\)
\(\cos^245^o=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}\)
=> \(A=1.44+\frac{1}{2}-\frac{1}{2}=44\)
=3tan67-3tan67+5[cos^2(16 độ)+cos^2(74 độ)]-1
=5-1
=4
a) \(cos^275+cos^253+cos^217+cos^237\)
ta áp dụng: \(sin^2a+cos^2a=1\)
ta được: \(\left(cos^275+cos^2\left(90-75\right)\right)+\left(cos^253+cos^2\left(90-53\right)\right)\)
=\(1+1=2\)
b) \(\frac{tan^215-1}{cot75-1}-cos75\)
=\(\frac{\left(tan15-1\right)\left(tan15+1\right)}{tan15-1}-cos75\)
=\(tan15+1-sin15\)=sin15\(\left(\frac{1}{cos15}-1+\frac{1}{sin15}\right)\)
a) \(cos^273^o+cos^253^o+cos^217^o+cos^237^o=\left(cos^273^o+cos^217^o\right)+\left(cos^253^o+cos^237^o\right)\)
\(=\left(cos^273^o+sin^273^o\right)+\left(cos^253^o+sin^253^o\right)=1+1=2\)
b) \(\frac{tan^215^o-1}{cotg75^o-1}-cos75^o=\frac{\left(tan15^o-1\right)\left(tan15^o+1\right)}{tan15^o-1}-cos75^o=tan15^o+1-cos75^o\)
Lời giải:
a)
\(A=\frac{\sin ^2a-\cos ^2a}{\sin a\cos a}=\frac{\sin a}{\cos a}-\frac{\cos a}{\sin a}=\frac{\sin a}{\cos a}-\frac{1}{\frac{\sin a}{\cos a}}=\tan a-\frac{1}{\tan a}\)
\(=\sqrt{3}-\frac{1}{\sqrt{3}}\)
b)
Sử dụng công thức: \(\sin ^2a+\cos ^2a=1; \cos a=\sin (90-a); \tan a=\cot (90-a)\) ta có:
\(B=\cos ^255^0-\cot 58^0+\frac{\tan 52^0}{\cot 38^0}+\cos ^235^0+\tan 32^0\)
\(=\sin ^2(90^0-55^0)-\tan (90^0-58^0)+\frac{\tan 52^0}{\tan (90^0-38^0)}+\cos ^235^0+\tan 32^0\)
\(=(\sin ^235^0+\cos ^235^0)-\tan 32^0+\tan 32^0+\frac{\tan 52^0}{\tan 52^0}\)
\(=1+0+1=2\)
Lời giải:
a)
\(A=\frac{\sin ^2a-\cos ^2a}{\sin a\cos a}=\frac{\sin a}{\cos a}-\frac{\cos a}{\sin a}=\frac{\sin a}{\cos a}-\frac{1}{\frac{\sin a}{\cos a}}=\tan a-\frac{1}{\tan a}\)
\(=\sqrt{3}-\frac{1}{\sqrt{3}}\)
b)
Sử dụng công thức: \(\sin ^2a+\cos ^2a=1; \cos a=\sin (90-a); \tan a=\cot (90-a)\) ta có:
\(B=\cos ^255^0-\cot 58^0+\frac{\tan 52^0}{\cot 38^0}+\cos ^235^0+\tan 32^0\)
\(=\sin ^2(90^0-55^0)-\tan (90^0-58^0)+\frac{\tan 52^0}{\tan (90^0-38^0)}+\cos ^235^0+\tan 32^0\)
\(=(\sin ^235^0+\cos ^235^0)-\tan 32^0+\tan 32^0+\frac{\tan 52^0}{\tan 52^0}\)
\(=1+0+1=2\)